$h(x) = 6x^{2}+3(f(x))$ $f(x) = 2x^{2}-4x$ $ f(h(1)) = {?} $
Explanation: First, let's solve for the value of the inner function, $h(1)$ . Then we'll know what to plug into the outer function. $h(1) = 6(1^{2})+3(f(1))$ To solve for the value of $h$ , we need to solve for the value of $f(1)$ $f(1) = 2(1^{2})+(-4)(1)$ $f(1) = -2$ That means $h(1) = 6(1^{2})+(3)(-2)$ $h(1) = 0$ Now we know that $h(1) = 0$ . Let's solve for $f(h(1))$ , which is $f(0)$ $f(0) = 2(0^{2})+(-4)(0)$ $f(0) = 0$